{"id":7815,"date":"2011-07-15T07:15:32","date_gmt":"2011-07-15T15:15:32","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/?p=7815"},"modified":"2011-07-08T08:55:24","modified_gmt":"2011-07-08T16:55:24","slug":"quick-mental-math-averages","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/quick-mental-math-averages\/","title":{"rendered":"Quick Mental Math: Averages"},"content":{"rendered":"

This post was written by Rich Zwelling. For more GMAT prep<\/a> tips, check out the Knewton GMAT blog<\/a>. <\/em><\/p>\n

I run the risk of sounding like a hypocrite here. I'm often fond of telling my students that the GMAT is a reasoning test, not a speed calculation test.\u00a0 (I say this to assuage their fears after I break the bad news that no calculators are allowed.\u00a0 I tell them that even if they had calculators, they wouldn't be of much help, because the arithmetic itself is pretty straightforward.\u00a0 It's the packaging of the questions and the required synthesis of information that make the test so difficult.)<\/p>\n

But if the GMAT is more about reasoning than speed calculations, then why am I writing a post on quick mental math?\u00a0 Well, it's true that any GMAT quant question can be solved without quick mental math (and certainly without a calculator).\u00a0 But that doesn't mean the test doesn't reward you if you can speed up the process.\u00a0 And by \"speed up,\" I don't mean cutting corners or being less rigorous.\u00a0 I mean recognizing that there are ways to make calculations easier and more manageable.<\/p>\n

Take averages.\u00a0 You're probably familiar with the basic way to find averages:\u00a0 you add up all the elements in a set of numbers, then divide by the number of terms, and the result is the average.\u00a0 For example, let's take the set (3, 7, 15, 9, 6).\u00a0 To calculate the average, we'd take 3+7+15+9+6 = 40, then divide that 40 by the number of elements (5) to get an average of 8.<\/p>\n

There's an important principle at work here:\u00a0 no matter what the values of those 5 elements are, they will always have an average of 8 if they sum to 40.\u00a0 Thus, if we change (3,7,15,9,6) to (4,6,15,9,6), the average does not change, because the numbers still sum to 40.\u00a0 This is shown in the \"rearranged average formula,\" which simply states that the sum of a group of numbers is equal to that group's average multiplied by the number of terms in the set (Sum = Avg * # terms).<\/p>\n

This leads to an incredibly helpful mental math trick.\u00a0 If you can change the individual numbers in the set such that they get closer to the average, while at the same time making sure the sum doesn't change, you can hone in on the exact average much much quicker.<\/p>\n

Let's use our original set of five numbers (3, 7, 15, 9, 6) to demonstrate:<\/p>\n

To start, let's take the average of individual pairs.\u00a0 For example, 3 and 7 have 5 in the middle, and 9 and 15 have 12 in the middle.\u00a0 We can therefore change the list from (3, 7, 15, 9, 6) to (5, 5, 12, 12, 6).\u00a0 Notice that the two sets still have the same average, because we've changed neither the sum of the numbers nor the number of elements in the set.<\/p>\n

We now have two 5s and two 12s. Let's break those down.\u00a0 The average of 5 and 12 is 8.5, so we can change (5, 5, 12, 12, 6) to (8.5, 8.5, 8.5, 8.5, 6).<\/p>\n

Obviously, the average of the set will not be 8.5, since you have that 6 thrown in. But on a Problem Solving question, you could possibly approximate; since 6 is not far off from 8.5, you know the average will be slightly less than 8.5, so if only one answer choice is slightly less than 8.5, you know it will be your winner.<\/p>\n

But for the more precise, let's finish what we started!\u00a0 Now comes another cool mental math trick related to averages.\u00a0 If all the elements were 8.5, then of course the average would be 8.5.\u00a0 But what effect does that 6 really have on the entire set? Well, it's as if we had all 8.5s and then reduced the entire sum of the set by 2.5 (since we changed one 8.5 to a 6).\u00a0 If we spread that 2.5 difference evenly across the entire set of five numbers, each number would be reduced by 2.5 \/ 5 = 0.5.\u00a0 Therefore, it's as if we had all 8.5s to start with and then reduced each one by 0.5.\u00a0 Guess what... that means the final average is 8.5 - 0.5 = 8.\u00a0 Done!<\/p>\n

Another way you can quickly find an average is to use a single number as a \"focal point.\"\u00a0 This works well if the numbers are somewhat close together, but far enough apart that the average might not be immediately obvious.\u00a0 For example, if I gave you the list (100, 95, 85, 110, 150) and asked you to find the average, you could pick 100 as your focus and record how much less or greater each other number is than 100:<\/p>\n

100 \u00a0\u00a0 95 \u00a0\u00a0\u00a0 85 \u00a0\u00a0\u00a0\u00a0110 \u00a0\u00a0\u00a0 150
\n(+0)\u00a0 (-5)\u00a0 (-15)\u00a0 (+10)\u00a0 (+50)<\/p>\n

Add up all the comparisons to get a net difference: \u00a0 -5 + -15 + 10 + 50 = +40.\u00a0 Again, this is the conceptual equivalent of having five 100s and adding 40 to the entire sum.\u00a0 So, if we spread that 40 evenly across the set, each term would increase by 40\/5 = 8 from the \"focal point\" of 100.\u00a0 And voil\u00e0<\/em>!\u00a0 That means the average is 100+8 = 108.<\/p>\n

Next week, I'll extend this to a discussion of weighted averages, which are very<\/em> common on the GMAT!<\/p>\n

In the meantime, try to apply these principles to the following official GMAT problem:<\/p>\n

Ada and Paul received their scores on three tests. On the first test, Ada's score was 10 points higher than Paul's score. On the second test, Ada's score was 4 points higher than Paul's score. If Paul's average (arithmetic mean) score on the three tests was 3 points higher than Ada's average score on the three tests, then Paul's score on the third test was how many points higher than Ada's score?<\/em><\/p>\n

A) 9 B) 14 C) 17 D) 23 E) 25<\/em><\/p>\n

* * *<\/p>\n

Answer to the problem from last week's post<\/a>:<\/strong><\/p>\n

Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce (5\/4)<\/em>w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?<\/em><\/p>\n

A) 4 B) 6 C) 8 D) 10 E) 12<\/em><\/p>\n

The first step is to realize that if it takes 3 days for the two machines to produce (5\/4)w<\/em> widgets, we can multiply both quantities by 4\/5 to find that it takes 12\/5 days to produce w<\/em> widgets. We do this because it's easier to think of the amount of work as simply w<\/em> widgets rather than (5\/4)w<\/em> widgets.<\/p>\n

Now, let's recall the formula we came up with a formula for combined time:<\/p>\n

If Machine X takes x days to complete the job on its own, and Machine Y takes y days to complete the job on its own, then the two machines working together at their constant rates would take (xy)\/(x+y) days to complete the job. If you'd like to review how we get that quantity, go back to last week's post<\/a>. According to what we just found, that quantity is 12\/5, so:<\/p>\n

(xy)\/(x+y) = 12\/5<\/p>\n

Now, here's where you can get REALLY clever! The prompt says that Machine X takes 2 days longer than Machine Y, so if you really wanted to, you could substitute x-2 for y in the equation above and solve for x.<\/p>\n

However, there's a very sneaky shortcut: Notice that we're trying to find the number of days it takes Machine X to produce 2w widgets. All the answer choices are even numbers, which means that the number of days it takes Machine X to produce w widgets (i.e. half of whatever the answer is) is an integer! So really, we just need to find integer values of x and y that are 2 apart and that satisfy the equation!<\/p>\n

(xy)\/(x+y) = 12\/5<\/p>\n

Since x+y is an integer, it will have to be a multiple of 5, according to the ratio. There's no way x+y could equal 5 if x and y are integer values 2 apart. You can show that with substitution or quick testing cases. But could x+y=10, in which case xy would have to be 24? Sure! x=6 and y=4 are numbers for which x is two greater than y and both x and y satisfy the equation! Therefore, it takes Machine X 6 days to produce w widgets, which means that it takes 12 days to produce 2w widgets.<\/p>\n

Final answer: E<\/p>\n","protected":false},"excerpt":{"rendered":"

This post was written by Rich Zwelling. For more GMAT prep tips, check out the Knewton GMAT blog. I run the risk of sounding like a hypocrite here. I’m often…<\/p>\n","protected":false},"author":104,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9,243,717,736],"tags":[],"class_list":["post-7815","post","type-post","status-publish","format-standard","hentry","category-gmat","category-blog","category-problem-solving-gmat","category-quant-gmat","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/7815","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/users\/104"}],"replies":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/comments?post=7815"}],"version-history":[{"count":1,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/7815\/revisions"}],"predecessor-version":[{"id":7816,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/7815\/revisions\/7816"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=7815"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=7815"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=7815"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}